What is the integral of sqrt(x^2-36) ?

The integral of sqrt(x^2-36) is 36(1/72 xsqrt(x^2-36)-1/2 ln((|x+sqrt(x^2-36)|)/6))+C

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integral of sqrt(x^2-36)

\int x^{2} - 36 d x

36 (\frac{1}{72} x x^{2} - 36 - \frac{1}{2} ln (\frac{x + x ^{2} - 36}{6})) + C

Solution steps

\int x^{2} - 36 d x

Apply Trigonometric Substitution

: \int 36 tan^{2} (u) sec (u) d u

= \int 36 tan^{2} (u) sec (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 36 \cdot \int tan^{2} (u) sec (u) d u

Rewrite using trig identities

= 36 \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

Expand

(- 1 + sec^{2} (u)) sec (u) : - sec (u) + sec^{3} (u)

= 36 \cdot \int - sec (u) + sec^{3} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 36 (- \int sec (u) d u + \int sec^{3} (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

= 36 (- ln ∣ tan (u) + sec (u) ∣ + \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arcsec (\frac{1}{6} x)

= 36 (- ln tan (arcsec (\frac{1}{6} x)) + sec (arcsec (\frac{1}{6} x)) + \frac{1}{2} sec (arcsec (\frac{1}{6} x)) tan (arcsec (\frac{1}{6} x)) + \frac{1}{2} ln tan (arcsec (\frac{1}{6} x)) + sec (arcsec (\frac{1}{6} x)))

Simplify

36 (- ln tan (arcsec (\frac{1}{6} x)) + sec (arcsec (\frac{1}{6} x)) + \frac{1}{2} sec (arcsec (\frac{1}{6} x)) tan (arcsec (\frac{1}{6} x)) + \frac{1}{2} ln tan (arcsec (\frac{1}{6} x)) + sec (arcsec (\frac{1}{6} x))) : 36 (\frac{1}{72} x x^{2} - 36 - \frac{1}{2} ln (\frac{x + x ^{2} - 36}{6}))

= 36 (\frac{1}{72} x x^{2} - 36 - \frac{1}{2} ln (\frac{x + x ^{2} - 36}{6}))

Add a constant to the solution

= 36 (\frac{1}{72} x x^{2} - 36 - \frac{1}{2} ln (\frac{x + x ^{2} - 36}{6})) + C

ln (\frac{1 - sin ( x )}{1 + sin ( x )})

\frac{d}{d x} (5^{2})

\frac{\partial}{\partial x} (r cos (θ))

\int_{1}^{6} \frac{5 ^{l n (x)}}{x} d x

n = 0 \sum \infty 3 (\frac{3}{2})^{n}

What is the integral of sqrt(x^2-36) ?
The integral of sqrt(x^2-36) is 36(1/72 xsqrt(x^2-36)-1/2 ln((|x+sqrt(x^2-36)|)/6))+C